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Câu hỏi: Cho số phức ${{z}_{1}},{{z}_{2}}$ thỏa mãn các điều kiện $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=2$ và $\left| {{z}_{1}}+4{{z}_{2}} \right|=8.$ Giá trị của $\left| 4{{z}_{1}}-{{z}_{2}} \right|$ bằng
A. $6\sqrt{2}$
B. $4\sqrt{2}$
C. $5\sqrt{2}$
D. 6
A. $6\sqrt{2}$
B. $4\sqrt{2}$
C. $5\sqrt{2}$
D. 6
Cách giải:
Đặt $z=\dfrac{{{z}_{1}}}{{{z}_{2}}}\Rightarrow \left| z \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{2}{2}=1.$
Ta có: $\left| z+4 \right|=\dfrac{\left| {{z}_{1}}+4{{z}_{2}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{8}{2}=4.$
Đặt $z=a+bi,$ khi đó ta có:
$\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1 \\
& {{\left( a+4 \right)}^{2}}+{{b}^{2}}=16 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a+4 \right)}^{2}}-{{a}^{2}}=15 \\
& {{a}^{2}}+{{b}^{2}}=1 \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& 8a+16=15 \\
& {{a}^{2}}+{{b}^{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{8} \\
& {{b}^{2}}=\dfrac{63}{64} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{8} \\
& b=\pm \dfrac{\sqrt{63}}{8} \\
\end{aligned} \right.$
$\Rightarrow z=-\dfrac{1}{8}\pm \dfrac{\sqrt{63}}{8}i$
$\Rightarrow \left| 4{{z}_{1}}-{{z}_{2}} \right|=\left| 4z.{{z}_{2}}-{{z}_{2}} \right|=\left| 4z-1 \right|.\left| {{z}_{2}} \right|$
Ta có $\left| 4z-1 \right|=\left| -\dfrac{1}{2}\pm \dfrac{\sqrt{63}}{2}i-1 \right|=\left| -\dfrac{3}{2}\pm \dfrac{\sqrt{63}}{2}i \right|=\sqrt{\dfrac{9}{4}+\dfrac{63}{4}}=3\sqrt{2}.$
Vậy $\left| 4{{z}_{1}}-{{z}_{2}} \right|=3\sqrt{2}.2=6\sqrt{2}.$
Đặt $z=\dfrac{{{z}_{1}}}{{{z}_{2}}}\Rightarrow \left| z \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{2}{2}=1.$
Ta có: $\left| z+4 \right|=\dfrac{\left| {{z}_{1}}+4{{z}_{2}} \right|}{\left| {{z}_{2}} \right|}=\dfrac{8}{2}=4.$
Đặt $z=a+bi,$ khi đó ta có:
$\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1 \\
& {{\left( a+4 \right)}^{2}}+{{b}^{2}}=16 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a+4 \right)}^{2}}-{{a}^{2}}=15 \\
& {{a}^{2}}+{{b}^{2}}=1 \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& 8a+16=15 \\
& {{a}^{2}}+{{b}^{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{8} \\
& {{b}^{2}}=\dfrac{63}{64} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{8} \\
& b=\pm \dfrac{\sqrt{63}}{8} \\
\end{aligned} \right.$
$\Rightarrow z=-\dfrac{1}{8}\pm \dfrac{\sqrt{63}}{8}i$
$\Rightarrow \left| 4{{z}_{1}}-{{z}_{2}} \right|=\left| 4z.{{z}_{2}}-{{z}_{2}} \right|=\left| 4z-1 \right|.\left| {{z}_{2}} \right|$
Ta có $\left| 4z-1 \right|=\left| -\dfrac{1}{2}\pm \dfrac{\sqrt{63}}{2}i-1 \right|=\left| -\dfrac{3}{2}\pm \dfrac{\sqrt{63}}{2}i \right|=\sqrt{\dfrac{9}{4}+\dfrac{63}{4}}=3\sqrt{2}.$
Vậy $\left| 4{{z}_{1}}-{{z}_{2}} \right|=3\sqrt{2}.2=6\sqrt{2}.$
Đáp án A.